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Mamont248 [21]
3 years ago
10

At the 2012 Summer Olympic Games in London, in the Men's Shot Put qualifying round, the distances ranged from 17.58 meters to 21

.36 meters with a mean distance of 19.64 meters and a standard deviation equal to 0.92 meters. If the distribution were truly a normal distribution, what percent of the athletes would have had a distance greater than 20.56 meters, one standard deviation above the mean? There were 40 athletes in the competition. How many athletes does that represent?
a. about 20 athletes
b. about 26 athletes
c. about 6 athletes
Mathematics
1 answer:
chubhunter [2.5K]3 years ago
8 0

Answer:

One standard deviation above the mean means a z-score of 1.0. If we put this into the calculator as normalcdf(1,999) = .159

15.9 percent of the athletes would have a distance greater than 20.56 meters.

Now multiply the number of athletes by the percent.

.159 * 40 = 6.346 = about 6 athletes

c. about 6 athletes

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A marina is in the shape of a coordinate grid. Boat A is docked at (3.1, −1) and Boat B is docked at (−4.1, −1). The boats are _
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<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

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Step-by-step explanation:

<u>Step 1: Define</u>

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Boat B(-4.1, -1)

<u>Step 2: Find distance </u><em><u>d</u></em>

Simply plug in the 2 coordinates into the distance formula to find distance <em>d</em>.

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